Find the Confidence Intervals

The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 13.2,  5.1,  7.5,  8.0,  12.7,  7.6,  13.8,  14.5,  7.7, and 10.5.  Determine a 95 percent confidence interval for the mean time for all players.

• 12.30 < μ < 7.82

•  7.82 < μ < 12.30

• 7.72 < μ < 12.40

• 12.40 < μ < 7.72

A 90% confidence interval for the mean speed driven by cars on Due West Road is given by (21, 45). What was the sample mean of this study?

• 12

• 24

• 33

• None of these.

A 90% confidence interval for the mean speed driven by cars on Due West Road is given by (21, 45). What was the margin of error of this interval?

• 12

• 24

• 33

• None of these.

A 90% confidence interval for the average salary of all CEOs in the electronics industry was constructed using the results of a random survey of 45 CEOs. The interval was $139,048 to $154,144. Give the interpretation of the interval.

• 90% of the sampled CEOs have salaries that fell in the interval $139,048 to $154,144

• We are 90% confident that the mean salary of all CEOs in the electronics industry falls in the interval $139,048 to $154,144.

• There is a 90% chance that CEOs in the electronics industry have salaries that fall between $139,048 to $154,144

• We are 90% confident that the mean salary of the sampled CEOs falls in the interval $139,048 to $154,144.

In a survey of 104 students, it was found that 79 went to the homecoming game this year. Calculate a 99% confidence interval for p.

• (0.652, 0.868)

• (0.691, 0.829)

• (0.678, 0.842(

• (0.685, 0.895)

In a survey of 104 students, it was found that 79 went to the homecoming game this year. A 99% confidence interval for p is (0.652, 0.868). Interpret this interval.

• 99% of the time the true proportion of people who went to the homecoming game this year is between 65.2% and 86.6%.

• The probability that the population proportion of people who went to the homecoming game this year is between 65.2% and 86.6% is 95%.

• Based on this sample, I am 99% confident that the true proportion of people who went to the homecoming game this year is between 65.2% and 86.6%.

• 99% of all possible intervals calculated this way will capture the true proportion of people who went to the homecoming game this year

Which confidence would result in a wider interval (assuming nothing else changed), 91% or 97%?

• 97%, z* is larger for higher confidence intervals

• 97%, the standard error is larger for higher confidence intervals

• 91%, z* is larger for lower confidence intervals

• 91%, the standard error is smaller for lower confidence intervals

For a sample, the 95% confidence interval is (0.202, 0.482). What is the point estimate, p̂, of this sample?

• 0.28

• 0.14

• 0.342

• 0.684

What is the margin of error of this sample?

• 0.28

• 0.14

• 0.342

• 0.684

If 827 represents "no" voters out of 2584, construct a 95% confidence interval for the "no" group.

• [.300,.340]

• [.298,.342]

• [.302,.338]

• [.288,.352]